Wire of length 72 m is divided into two pieces and the pieces are bent into a square and a circle. How many meters of wire should be allotted to the circle so that the sum of the areas is minimized? Find a positive number x such that the sum of 2sx and possible. Does this problem require optimization over an open interval or a closed interval?

For the circle, it should be alloted 31.673 meters of wire. For common sense, this problem requires optimization over an open interval (you shouldnt count the borders as possibilities, since you may expect both the square and the circle to have positive perimeter).

Step-by-step explanation:

Lets late x m for the perimeter of the circle.

The perimeter of a circle is 2πr, therefore

x = 2πr

r = x/2π

The area of the circle is πr² = π(x/2π)² = x²/4π

If we take x for the circle then we have 72-x for the square. The perimeter of a square is 4 times the length of a side, as a result, the side length is (72-x)/4 = 18 - x/4.

The area of the square is (18 - x/4)² = 324-9x+x²/16

Lets call f(x) the sum of the areas,

f(x) = x² ( 1/4π + 1/16) -9x + 324. Note that f is a quadratic function with positive main coefficient, hence its minimum is reached in its vertex.

The vertex of x is found in the point

-(-9) / 2(1/4π + 1/16) = 31.673

Therefore, a value x=31.673 minimizes the sum of the areas.

You could use either an open interval or a closed one. For common sense, it should be a open interval because you cant create neither a circle or a square of perimeter 0, but with an open interval, there migth not neccesarily be a minimum all the time. However, I would say that an open interval is more preferred than a closed one.